anonymous
  • anonymous
Important Question. Please explain with steps.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
@Mani_Jha , @saifoo.khan , @heena , @supercrazy92 please help.
saifoo.khan
  • saifoo.khan
No chem! :(

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anonymous
  • anonymous
Try at least.
saifoo.khan
  • saifoo.khan
@Mani_Jha is a boss at chem.
anonymous
  • anonymous
Yeah, that's why i called Mani Bhai also. He is my teacher. I hope he will do it this time too.:)))))
Mani_Jha
  • Mani_Jha
Let the concentration of NH3 be x. Now, the resultant solution was basic(as it had to be neutralized by NaOH). During the reaction of neutralization: M1V1=M2V2 x20=30*0.1 Find z. Let the initial volume of NH3 be V. Then when NH3 reacted initially with H2S04, 100 ml of H2S04 neutralized (v-20) ml of NH3(Because 20 ml was un-neutralized and made the solution basic) Again use the law of equivalents in this case(be careful: the n-factor of H2S04 is 2) Did you understand? @Saifoo.khan, you just guessed that, didnt u?
anonymous
  • anonymous
Mani Bhai, can u show the second equation what we need to take?
Mani_Jha
  • Mani_Jha
(v-20)x=100*0.1*2. ok?
anonymous
  • anonymous
Okay, mani bhaiiiiiii
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Wordings of these problems are terrible; teachers should bear in mind that students are not reading their minds and that wordings should be unambiguous. In first problem, I find 17 millimole NH3 or 0.38 litre in STP. That holds if "equivalence" in the wording is meant for 'first equivalence', because eventually, NH4+ formed will also be titrated. In second problem there should be 1.5 millimole of carbonate and 0.5 millimole of hydrogencarbonate. Here again, problem should read: "The second end point was reached when AN EXTRA 20 ml of the acid was added
anonymous
  • anonymous
Oooookkkkkk, I will consult with my teacher about this. But these r the exact wordings that our teacher gave.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
What about the answers?
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
btw, do you use 22.4 litres/mole or 22.7 litres/mole for molar volume in STP?
anonymous
  • anonymous
22.4
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Do you have expected answers ?
anonymous
  • anonymous
Not now. i need to submit it as my assignment. i couldn't do it, so i asked it here.
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
Do my answers coincide with yours of Mani-Jha's? Do you understand what reactions are at stake here?
anonymous
  • anonymous
My teacher gave the answer just now as 112 ml for the first one, and for the second one: Weight of NaOH = 0.8g, weight of Na2CO3 = 1.59 g
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
1st one is correct. I forgot to divide something by 5, as you take only a 20 mL sample of the 100 mL solution. So there were 5 millimoles of ammonia ie a volume of 112 mL
anonymous
  • anonymous
Thanks a lot
Vincent-Lyon.Fr
  • Vincent-Lyon.Fr
I have 159 mg of Na2CO3 which is not 1.59 g, unless concentration af acid is 1M instead of 0.1M. Then I do not understand the NaOH part. Your solution is supposed to have NaHCO3. What do you think?
anonymous
  • anonymous
Even I think the same.

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